Question Number 127888 by A8;15: last updated on 02/Jan/21
Answered by mindispower last updated on 02/Jan/21
$${x}={sh}\left({t}\right) \\ $$$$\Rightarrow{dx}={ch}\left({t}\right){dt} \\ $$$$\Leftrightarrow\int_{\mathrm{0}} ^{\infty} {sin}\left({ch}\left({t}\right)\right){cos}\left({sh}\left({t}\right)\right){dt}..=\Omega \\ $$$$\Lambda=\int_{\mathrm{0}} ^{\infty} {cos}\left({ch}\left({t}\right)\right){sin}\left({sh}\left({t}\right)\right){dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} {sin}\left({e}^{{t}} \right){dt}=\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}}{dx}=\frac{\pi}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\infty} {sin}\left({ch}\left({t}\right)+{sh}\left({t}\right)\right){dt}=\int_{\mathrm{0}} ^{\infty} {sin}\left({e}^{{t}} \right){dt}=\int_{\mathrm{1}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}}{dx} \\ $$$$\Delta=\int_{\mathrm{0}} ^{\infty} \left({sin}\left({ch}\left({t}\right)\right){cos}\left({sh}\left({t}\right)\right)+{cos}\left({ch}\left({t}\right)\right){sin}\left({sh}\left({t}\right)\right)\right){dt} \\ $$$$\Omega+\Lambda=\int_{\mathrm{1}} ^{\infty} \frac{{sin}\left({x}\right)}{{t}}{dt} \\ $$$$\Omega−\Lambda=\int_{\mathrm{0}} ^{\infty} {sin}\left({e}^{−{t}} \right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left({t}\right)}{{t}}{dt} \\ $$$$\Rightarrow\mathrm{2}\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left({t}\right)}{{t}}{dt}+\int_{\mathrm{1}} ^{\infty} \frac{{sin}\left({t}\right)}{{t}}{dt}=\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({t}\right)}{{t}}{dt}=\frac{\pi}{\mathrm{2}} \\ $$$$\Leftrightarrow\Omega=\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by A8;15: last updated on 03/Jan/21
$$\mathrm{thanks}\:\mathrm{sir}.\:\boldsymbol{\mathrm{Happy}}\:\boldsymbol{\mathrm{New}}\:\boldsymbol{\mathrm{Year}} \\ $$
Commented by mindispower last updated on 07/Jan/21
$${withe}\:{pleasur} \\ $$